Q.Write the structures of the isomers of alcohols with molecular formula $C_4H_{10}O$. Which one of these exhibits optical activity?

The key idea is that alcohols with formula $C_4H_{10}O$ are saturated, monohydric alcohols — four carbon atoms in a chain or branched, with an $-$OH group.

Reasoning:

1. Draw all possible carbon skeletons for 4 carbons: straight chain (butane) and branched (isobutane).
2. Place the $-$OH group on distinct carbons for each skeleton, avoiding duplicates.
3. Check for a chiral carbon (carbon with four different groups) — only one isomer has this.

The four structural isomers are:

• Butan-1-ol: $CH_3CH_2CH_2CH_2OH$
• Butan-2-ol: $CH_3CH_2CH(OH)CH_3$ …

The molecular formula $\mathrm{C_4H_{10}O}$ fits the general formula $\mathrm{C_nH_{2n+2}O}$ for a saturated alcohol or ether. Since we are asked for alcohols, the functional group is $-\mathrm{OH}$ attached to a carbon chain. The key to finding all isomers is to vary the carbon skeleton (straight vs. branched) and the position of the $-\mathrm{OH}$ group.

Optical activity arises when a molecule has no plane of symmetry — most commonly because it contains a carbon atom bonded to four different groups. That carbon is called a chiral centre (or stereocentre). For a molecule to be optically active, it must exist as non-superimposable mirror images (enantiomers). So among the isomers, we look for one with a chiral carbon.

Let’s build the isomers systematically.

1. Straight-chain (n-butane) skeleton: $\mathrm{C-C-C-C}$

   Place the $-\mathrm{OH}$ at the end of the chain: $\mathrm{CH_3CH_2CH_2CH_2OH}$ — this is butan-1-ol (a primary alcohol).

   Place the $-\mathrm{OH}$ on the second carbon: $\mathrm{CH_3CH_2CH(OH)CH_3}$ — this is butan-2-ol (a secondary alcohol).

   No other positions are possible on a four-carbon straight chain (positions 3 and 4 are identical to 2 and 1 by symmetry).

2. Branched skeleton (isobutane): $\mathrm{C-C(C)-C}$

   The carbon backbone is $\mathrm{CH_3CH(CH_3)CH_3}$ (2-methylpropane).

   Place $-\mathrm{OH}$ on a terminal carbon: $\mathrm{(CH_3)_2CHCH_2OH}$ — this is 2-methylpropan-1-ol (a primary alcohol).

   Place $-\mathrm{OH}$ on the central (tertiary) carbon: $\mathrm{(CH_3)_3COH}$ — this is 2-methylpropan-2-ol (a tertiary alcohol).

   No other distinct positions exist.

So we have exactly four structural isomers:

Now, which one is optically active? Check each for a chiral carbon.

• Butan-1-ol: Carbon-2 has two H atoms, carbon-3 has two H atoms — no carbon is bonded to four different groups. No chirality.
• 2-Methylpropan-1-ol: The carbon bearing $-\mathrm{OH}$ is $\mathrm{CH_2OH}$ (two H atoms). The central carbon is $\mathrm{CH}$ bonded to two identical methyl groups — not chiral. …

Method: Systematic Isomer Enumeration + Optical Activity Check

Step 1 – Identify the functional group and degree of unsaturation

The formula is $C_4H_{10}O$.

For a saturated acyclic compound, the formula would be $C_nH_{2n+2} = C_4H_{10}$.

Here we have one oxygen — alcohols ($-OH$) are saturated, so no double bonds or rings.

We are looking for all structural isomers of alcohols.

Step 2 – Draw all carbon skeletons for 4 carbons

Two possible skeletons:

• Straight chain: $C-C-C-C$ (butane skeleton)
• Branched chain: $C-C(C)-C$ (isobutane skeleton)

Step 3 – Place the $-OH$ group on distinct carbons

From straight chain (butane):

1. Butan-1-ol (primary alcohol)

   $CH_3-CH_2-CH_2-CH_2OH$

2. Butan-2-ol (secondary alcohol)

   $CH_3-CH(OH)-CH_2-CH_3$

From branched chain (isobutane):

3. 2-Methylpropan-1-ol (primary alcohol)

   $(CH_3)_2CH-CH_2OH$

4. 2-Methylpropan-2-ol (tertiary alcohol)

   $(CH_3)_3C-OH$

Step 4 – Check for optical activity

Optical activity requires a chiral carbon — a carbon with four different substituents.

• Butan-1-ol: No chiral carbon (terminal $-OH$ group). …

Step 1: Understand the formula

The molecular formula is $C_4H_{10}O$.

For an alcohol ($R-OH$), the general formula is $C_nH_{2n+2}O$.

Here $n=4$, so $C_4H_{10}O$ fits perfectly — no unsaturation, only saturated alcohols (and ethers, but we focus on alcohols here).

Step 2: Draw all alcohol isomers

We need to arrange 4 carbons in a chain with an –OH group attached. The key: change the carbon skeleton and position of –OH.

1. Butan-1-ol (primary alcohol)

CH3-CH2-CH2-CH2-OH

• Straight chain, –OH at terminal carbon.

2. Butan-2-ol (secondary alcohol)

CH3-CH2-CH(OH)-CH3

• Straight chain, –OH at carbon #2.

3. 2-Methylpropan-1-ol (primary alcohol)

CH3-CH(CH3)-CH2-OH

• Branched chain (isobutyl group), –OH at terminal carbon.

4. 2-Methylpropan-2-ol (tertiary alcohol)

CH3-C(OH)(CH3)-CH3

• Branched chain, –OH at the central (tertiary) carbon.

Total alcohol isomers = 4

Step 3: Which one exhibits optical activity?

Optical activity requires a chiral carbon — a carbon with four different substituents.

• Butan-1-ol: No chiral carbon (all carbons have at least two identical H’s or groups).
• Butan-2-ol: Carbon #2 has –OH, –H, –CH$_3$, –CH$_2$CH$_3$ → four different groups → chiral → exhibits optical activity.
• 2-Methylpropan-1-ol: No chiral carbon.
• 2-Methylpropan-2-ol: Central carbon has two identical –CH$_3$ groups → not chiral.

Answer: Butan-2-ol shows optical activity.

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